# Lagrangian of a disk with a hole on an inclined plane

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1. Mar 20, 2016

### Physgeek64

1. The problem statement, all variables and given/known data
A wheel consists of a circular uniform disk with a circular hole in it. The disc is
of radius R and mass per unit area ρ. The hole is of radius ro and an axle of radius ro
passes through it. The centre of the hole is offset radially from the centre of the disk by
ro. The wheel rotates without friction about the axle. Using the parallel axis theorem,
or otherwise, evaluate an expression for the moment of inertia of the wheel around the
axle. The angular speed of the wheel is ω, what is its kinetic energy?

The wheel is removed from the axle and rolls without slipping along a flat surface with the same angular speed. Show that its rotational kinetic energy is less than for the case of rotation about the axle

The wheel is placed on a surface sloping at an angle α where it rolls without slipping. It starts from rest with the centre of the hole directly above the centre of the disk. Neglecting the contribution of the hole to kinetic energy, use the Euler Lagrange method to find the equation of motion

2. Relevant equations
L=T-V
KE of rotation= 1/2*I*(ω)^2

(Most are below)

3. The attempt at a solution
So I think I've done the first two parts okay- I got the the moment of inertia for the first part to be 1/2*pi*rho*(R^4+R^2*r^2-r^4) and the KE to be half that times the ω^2

For the next part I for the KE to be 1/2*pi*rho*ω^2*(R^4-3r^4), and showed that the condition for the KE to be less is R>sqrt(2)r, which is obviously satisfied by the geometry of the system

For the last part, Im a bit confused because the centre of mass does not move with in a straight path, and is rotating

So I first found the location of the COM relative to the centre of the disc, and got this to be -r^3/(R^2-r^2)

So the rational KE of the COM is 1/2*((pi*rho)*(r^6)*ω^2)/((R^2-r^2) (sorry- Im still not sure how to use latex properly! I will be practicing, so please bare with me) Also- Here ω is varying- but I thought this would be neater to write than 'theta dot'

The rotational KE of the disc, ignoring the hole is

1/4*(pi*rho)*ω^2*R^4

But then Im not sure what the translational KE of the disc is since the COM is moving downwards and rotating, so its translational KE is modulating. I'm not sure if the question permits me to simply ignore the hole completely and simply use the rational KE of the disc as if it were solid, and the translational KE of the COM as if the disc were solid. If so then I think I can almost do this question.
And similarly, for the PE since the height is moving downwards, but periodically varying as it rotates I'm struggling to see how to formulate it

Edit:
So far for the potential energy I have

PE= -M*g*s*sin(alpha) - Mg*(centre off mass position)*sin(theta), where theta is the integral of ω and M=pi*rho*(R^2-r^2),
but Im not sure if this is correct?

Thanks in advance :)

Last edited: Mar 20, 2016
2. Mar 20, 2016

### Staff: Mentor

I don't see how you got this (would it vanish for r->0?) and I also don't think it is necessary. It is easier to work in the frame of the geometric center of the disk.